The solubility of silver bromide in hypo solution (excess) is due to the formation of:

Options

A. `Ag_(2)SO_(3)`
B. `Ag_(2)S_(2)O_(3)`
C. `[Ag(S_(2)O_(3))]^(-)`
D. `[Ag(S_(2)O_(3))_(2)]^(3-)`

Answer

To solve the question regarding the solubility of silver bromide (AgBr) in hypo solution (sodium thiosulfate, Na2S2O3), we can follow these steps:

### Step 1: Understand the Context
Silver bromide (AgBr) is a sparingly soluble salt that can dissolve in certain solutions. In this case, we are looking at its solubility in hypo solution, which is sodium thiosulfate.

**Hint:** Consider the properties of silver bromide and what happens when it interacts with sodium thiosulfate.

### Step 2: Write the Reaction
When silver bromide is treated with sodium thiosulfate, a reaction occurs. The sodium thiosulfate acts as a complexing agent. The reaction can be written as:

\[ \text{AgBr} + 2 \text{Na}_2\text{S}_2\text{O}_3 \rightarrow \text{Na}_3\text{AgS}_2\text{O}_3 + \text{NaBr} \]

**Hint:** Identify the products formed in the reaction and recognize the role of sodium thiosulfate.

### Step 3: Identify the Products
From the reaction, we see that the products formed are:
1. Sodium thiosulfate silver complex: \(\text{Na}_3\text{AgS}_2\text{O}_3\)
2. Sodium bromide: \(\text{NaBr}\)

The complex ion \(\text{AgS}_2\text{O}_3^{3-}\) is responsible for the increased solubility of silver bromide in the hypo solution.

**Hint:** Focus on the complex ion formed and its charge.

### Step 4: Conclusion
The solubility of silver bromide in excess hypo solution is due to the formation of the complex ion \(\text{AgS}_2\text{O}_3^{3-}\), which is part of the product \(\text{Na}_3\text{AgS}_2\text{O}_3\).

**Hint:** Review the options provided in the question to confirm which one corresponds to the complex ion formed.

### Final Answer
The correct answer is: **\(\text{AgS}_2\text{O}_3^{3-}\)** (the complex ion formed).